Question

The moment of inertia of a uniform solid sphere about its diameter is $I$. Its moment of inertia about a tangent is __________.
$\text A. \quad \dfrac{4}{3} I$
$\text B. \quad \dfrac{5}{4} I$
$\text C. \quad \dfrac{5}{3} I$
$\text D. \quad \dfrac{7}{2} I$

Answer 1

Hint: To find the moment of inertia of any shape about an unknown axis, first we need to know the moment of inertia of the shape about an axis either perpendicular to it or parallel to it. If we know the moment of inertia of a shape about an axis passing through the centre of mass of the body, then we can find the moment of inertia about any axis parallel to it by the use of parallel axis theorem.

Formula used:
${ I }_{ axis }={ I }_{ com }\ +\ Md^{ 2 }$ [ Statement of parallel axis theorem ]
Where M is the mass of the body and d is the distance of the axis from the centre of mass ( COM ) of the body.

Complete step by step answer:
Since every diameter of a sphere passes through its centre of mass, hence the moment of inertia of a sphere about its diameter is also the moment of inertia about its centre of mass. Now, to find the moment of inertia about its tangent, the distance between diameter and a tangent equals ‘R’.
Given, $I_{com} = I$ and $d= R$

Now using parallel axis theorem:
$I_{axis} = I + MR^2$ ------ 1.
As we know, the moment of inertia of solid sphere about its an axis passing through its centre of mass
is $\dfrac{2}{5}MR^2$.
Hence $I$=$\dfrac{2}{5}MR^2$
Or $MR^2 = \dfrac{5}{2}I$
Now putting $MR^2$ in equation 1. we get,
$I_{axis} = I + \dfrac{5}{2}I= \dfrac{7}{2}I$

So, the correct answer is “Option D”.

 Note:
While using the Parallel axis theorem, one axis should always be the one passing through the centre of mass. Chance of mistakes is high if one avoids this information. Students are advised to learn the moment of inertia of commonly used shapes for example: Disc, sphere, etc. $\dfrac25MR^2$ is the moment of inertia of a solid sphere about an axis passing through its centre of mass.